Hi a question about supporting electrolytes (SE) I hope you can help with. I found some related questions here through searching but looking for a simple answer (if there is one!):

If SE are electroinactive (in the potential range of interest) and so do not pass or receive electrons at the electrodes to complete the circuit then how to they "increase conductivity"? I am referring to typical DC experiments, e.g. chronoamperometry. It's not like electrons hop through the solution from one electrode to the other using SE ions as a shuttle.

I understand that one role is that they migrate to the opposite electrodes to neutralize ion charge imbalances that would build up due to reagent consumption or product generation from the electroactive species you are measuring (e.g oxidizing an enzyme product), thereby facilitating these redox reactions that do occur to continue. Is this the answer? So in just water + SE, for example, it is more conductive because they facilitate the electrolysis of water (H2 and O2 generation (e.g for SE without Cl-) by this method?

I also understand that movement of ions is considered current, but with DC wouldn't this migration slow down and exhaust as time went on? Is it this transient conductivity due to ion movement that they refer to?


Thanks for your answer Maurice. What you describe sounds like this (maybe not clearly written) part of my own possible answer to my question: "I understand that one role is that they migrate to the opposite electrodes to neutralize ion charge imbalances that would build up due to reagent consumption or product generation from the electroactive species you are measuring, thereby facilitating these redox reactions that do occur to continue".

I think my confusion arises because the IUPAC definition is: "A supporting electrolyte, in electrochemistry is an electrolyte containing chemical species that are not electroactive (within the range of potentials used) and which has an ionic strength and conductivity much larger than those due to the electroactive species added to the electrolyte", which makes it sound like the the SE ions themselves are conductive. Plus the most common explanation offered is along the lines of "because they are ions and ions conduct".

If I understand correctly now as discussed, they actually facilitate the reaction of the true electroactive species (which produce faradaic currents through redox - the current being a measure of conductivity), by maintaining neutrality around the electrode, rather than being "conductive" themselves (along with other roles such as making results more purely diffusion controlled, etc).

Put simply, the SE ions do not increase conductivity directly by contributing to the current (like an electroactive ion would), but increase conductivity by increasing the rate of faradaic processes of the actual electroactive species that are present (by using the SE ions to maintain electrical neutrality around the electrode).

Is this it?

Again, thanks for your time!

  • $\begingroup$ You are mixing many things. Start from here youtube.com/watch?v=QiCSTlgFwCY and perhaps refine your query. $\endgroup$
    – AChem
    Aug 24, 2020 at 1:53
  • 1
    $\begingroup$ Thank you for you attention to my question. That video discusses the role of SE in making the results being controlled by diffusion, rather than migration and diffusion, which i understand OK. My question is how electroinactive ions of the SE "increase the conductivity" of the solution, as is commonly stated, (when if they are electroinactive they do not produce any (Faradaic) current themselves). $\endgroup$
    – Bob Tomas
    Aug 24, 2020 at 16:22
  • $\begingroup$ @ Bob Tomas. Your supplementary information is correct. One might add that the negative pole (cathode) attracts positive ions (cations) and that the positive pole (anode) attracts negative ions. The movements of these ions create the electric currant in solution. $\endgroup$
    – Maurice
    Aug 24, 2020 at 19:07
  • $\begingroup$ Thanks for the followup and informative answer! I would accept your answer too if I knew how to accept more than one! $\endgroup$
    – Bob Tomas
    Aug 24, 2020 at 19:10

2 Answers 2


Remember that in electrochemistry whenever a direct current is passing through the solution there must be electrolysis occurring at the electrode. Now, all you want to see in electrochemical experiment, is current due to the oxidation or the reduction of the desired solute which is diffusing to the electrode in an unstirred cell.

The gist of your excellent and thought provoking question is:

My question is how electroinactive ions of the SE "increase the conductivity" of the solution, as is commonly stated, (when if they are electroinactive they do not produce any (Faradaic) current themselves).

After giving it a little thought and some literature search, it is clear that supporting electrolytes increase conductivity is an erroneous conclusion which is rampant everywhere. In the electrochemical experiment, the flux of the supporting electrolyte must be zero, it should not be carrying any current, because by definition, it is non-reactive in that potential window. The mass transport of the electro-inactive species must be zero.

*Yes, separately, if we were to measure the conductivity of a solution to which an supporting electrolyte, the conductivity would be higher than the pure solvent..but that is a separate measurement. Remember that conductivity is measured by very high frequency AC current.

The book, Electrochemical Engineering by Thomas F. Fuller, John N. Harb, explains it better, pg 69:

It is frequently reported or implied that the role of the supporting electrolyte is to reduce the electric field by carrying the current. This conclusion is in error since, at steady state, the flux of any nonreacting species must be zero, and the supporting electrolyte is by definition non- reacting. Therefore, only the fluxes of the reacting species contribute to the current (see Equation 4.4). That said, the presence of the supporting electrolyte does diminish the magnitude of the potential gradient. To explore this further, consider a system at steady state with no bulk flow where any transport of the supporting electrolyte due to migration must be offset by diffusion. Concentration gradients of the nonreacting species must be connected to those of the minor reacting species in order to preserve electroneutrality. Therefore, concentration gradients of the supporting species cannot be large. Furthermore, the conductivity of the supporting electrolyte is high. These two facts together make it impossible to support a large potential gradient in solution. As a result, migration can be neglected for the reacting species when its concentration is significantly less than that of the supporting electrolyte.

  • $\begingroup$ Many, many thanks! This has been bugging me for ages and many hours of searching. Finally, you say "the conductivity would be higher [with SE] than the pure solvent..but that is a separate measurement". Is there a simple explanation of what process is being measured in this other AC conductivity measurement (that is increased by SE)? $\endgroup$
    – Bob Tomas
    Aug 24, 2020 at 18:37
  • $\begingroup$ In those experiments, one is measuring the resistance of the solution by using AC current. The purpose of the AC current is to "undo" the electrolysis- remember whenever direct current is flowing in a solution, there must be electrolysis. AC trick is to avoid it the by very quickly changing the electrode polarity. You can accept the answer if you want. $\endgroup$
    – AChem
    Aug 24, 2020 at 19:04
  • $\begingroup$ Great! again thanks for taking the time to explain this. $\endgroup$
    – Bob Tomas
    Aug 24, 2020 at 19:12

It is not easy to understand what you are looking for. Maybe an example could help. Suppose you are using a solution of sodium sulfate $\ce{Na2SO4}$. The corresponding ions $\ce{Na+}$ and $\ce{SO4^{2-}}$are not discharged at the electrodes. On the contrary $\ce{H2O}$ is reacting at both electrodes. Let's start with the cathode.

At the cathode, $\ce{H2O}$ reacts with electrons coming from the outer cell, and produces $\ce{H2}$ and $\ce{OH-}$ ions. But this reaction can only happen if there is enough positive ions around the cathode to realize the electric neutrality of the solution. That is where $\ce{Na+}$ intervenes. Forget about $\ce{H2}$. If there is enough $\ce{Na+}$ at the cathode, $\ce{H2O}$ can react and produce $\ce{OH-}$ ions in solution. The rate of this reaction is proportional to the concentration of $\ce{Na+}$ ions in solution. This is why the presence of the electrolyte increases the reaction rate. Simultaneously, the negative ions $\ce{SO4^{2-}}$ are repelled by the cathode and attracted by the anode.

At the anode, $\ce{H2O}$ reacts in a similar manner to produce electrons. To do it, $\ce{H2O}$ is decomposed and produces $\ce{H+}$ ions in solution (or $\ce{H3O+}$) and $\ce{O2}$ gas. But this production of $\ce{H+}$ ions can only happen if there are enough negative ions in solution near the anode. If there are no negative ions, the first $\ce{H+}$ ions repell the next ones, and the reaction stops. This is where the sulfate ions intervene. The $\ce{H+}$ ions attract the $\ce{SO4^{2-}}$, and so the neutrality of the solution is maintained. The rate of the reaction is proportional to the concentration of the sulfate ions.

As a consequence, there is a general movement of positive ions towards the anode, and of negative ions towards the cathode. The amount of gases produced at the electrode is proportional to the concentration of electrolyte in solution.

  • $\begingroup$ Thanks Maurice, my question has been updated to respond to you (please see above). $\endgroup$
    – Bob Tomas
    Aug 24, 2020 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.